ii. What mass in grams of sodium trioxocarbonate (IV) is needed to prepare 250 cm³ of 0.2 moldm³ solution? [Na = 23, O=16, C=12]

here are the solutions to the chemistry questions: 4. (a) i. What is a buffer solution? A buffer solution is a solution that resists significant changes in pH when small amounts of an acid or a base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. 4. (a) ii. Give two uses of buffer solution. • Maintaining the pH of blood and other biological fluids. • Controlling the pH in industrial processes, such as fermentation and dyeing. 4. (b) i. Name the process used in the production of ammonia. • Haber process 4. (b) ii. Give two physical properties of nitrogen. • It is a colorless gas. • It is an odorless gas. 4. (b) iii. List the two allotropes of carbon. • Diamond • Graphite 4. (b) iv. Mention the structure of each allotrope in (b)iii. • Diamond: Each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement, forming a giant covalent network structure. • Graphite: Each carbon atom is covalently bonded to three other carbon atoms in a trigonal planar arrangement, forming hexagonal layers. These layers are held together by weak intermolecular forces. 4. (b) v. Give one use of charcoal. • Used as a fuel. 4. (c) i. Give the general molecular formula for: I. Alkenes: C_nH_2n II. Alkynes: C_nH_2n-2 4. (c) ii. Write out the possible structural isomers of C_4H_10. I. CH_3CH_2CH_2CH_3 II. CH_3CH(CH_3)CH_3 4. (c) iii. Give the IUPAC names of the isomers in (c)ii. I. Butane II. 2-Methylpropane 4. (c) iv. Calculate the molecular formula of a compound with the following composition by mass; 60% carbon, 13.3% hydrogen and 20.7% oxygen with a molar mass of 60. Note: The given percentages (60% C, 13.3% H, 20.7% O) sum to 94%. For the problem to be consistent with a molar mass of 60 g/mol, we assume the oxygen percentage is 100\% - 60\% - 13.3\% = 26.7\%. Step 1: Determine the mass of each element in 100 g of the compound. Mass of C = 60 g Mass of H = 13.3 g Mass of O = 26.7 g (assumed) Step 2: Convert the mass of each element to moles. Moles of C = 60 g12 g/mol = 5 mol Moles of H = 13.3 g1 g/mol = 13.3 mol Moles of O = 26.7 g16 g/mol ≈ 1.66875 mol Step 3: Determine the simplest whole-number mole ratio by dividing by the smallest number of moles. The smallest number of moles is 1.66875 (for Oxygen). Ratio C = (5)/(1.66875) ≈ 2.996 ≈ 3 Ratio H = (13.3)/(1.66875) ≈ 7.97 ≈ 8 Ratio O = (1.66875)/(1.66875) = 1 The empirical formula is C_3H_8O. Step 4: Calculate the empirical formula mass. Empirical formula mass = (3 × 12) + (8 × 1) + (1 × 16) = 36 + 8 + 16 = 60 g/mol Step 5: Determine the molecular formula. The given molar mass is 60 g/mol. Since the empirical formula mass (60 g/mol) is equal to the molar mass (60 g/mol), the molecular formula is the same as the empirical formula. The molecular formula is C_3H_8O. Send me the next one 📸